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Unformatted text preview: EE290T/BIOE265, Spring 2010 Principles of MRI Miki Lustig HW2 Solution 1. Nishimura 3.3 Answer 1 2 3 2. Precession in Tops (Double Points: brain teasers) (a) Explain why only spinning tops precess. That is, explain why static tops will not. Answer (b) Suppose you wanted to create an approximate macro version of the precessing quantum nuclear magnetic moment. Dr. Bore thinks you could do this by embedding a permanent magnetic (PM) moment within a toy top. Would you embed the PM perpendicular or parallel to the 4 rotation axis? If you altered the applied external magnetic eld, do you believe you would alter the mechanical precession frequency of the top that is, the precession would change from that determined merely by gravity? Could you create both positive and negative gyromagnetic ratios? 5 3. Magnetic eld steps Suppose you placed two test tubes (each 50 mL) at 1.5 T and one test tube (50 mL) at 1.6 T. Suppose both coils were visible with equal sensitivity to the RF coil. Sketch the intensity of the received signal as a function of frequency. Answer 6 4. RF Field (a) Find the amplitude of an RF pulse that performs a 90 degree excitation in exactly 1 ms at 1.5T. (b) Find the amplitude of an RF pulse that performs a 90 degree excitation in exactly 1 ms at 3T. Answer 7 5. Gradient eld bandwidth Sketch the spectrum of received signal on a 1.5 T MRI scanner versus frequency for an object of diameter = 25 cm with a gradient eld of peak ampitude 4 G/cm. Answer 8 6. RF Excitation. If we apply an RF waveform in addition to a gradient G, we can excite a slice through an object. Assume that the RF envelope B 1 ( t ) is the 6 ms segment of a sinc(), as shown 1 2 3 4 5 6 1 t, ms Write an expression for B 1 ( t ) , and its Fourier transform. Assuming a gradient G of 0.94 G/cm, how wide is the excited slice? Answer The RF waveform may be written B 1 ( t ) = sinc (2( t 3)) rect (( t 3) / 6) where times are in ms. The Fourier transform is F{ B 1 ( t ) } = e j 2 π 3 f 1 2 rect ( f/ 2) * 6 sinc (6 f ) The phase term doesn't e ect the slice width, so we'll ignore it. The Fourier transform is a 2 kHz wide rect convolved with a 1/6 kHz wide sinc. The result will be about 2 kHz wide. This is illustrated below, where the normalized functions are plotted, − 4 − 2 2 4 − 0.5 0.5 1 1.5 f, kHz rect(f/2) − 4 − 2 2 4 − 0.5 0.5 1 1.5 f, kHz sinc(6f) − 4 − 2 2 4 − 0.5 0.5 1 1.5 f, kHz rect(f/2)*6sinc(6f) To nd the width of the slice this pulse excites, rst note that a gradient of 0.94 G/cm produces aTo nd the width of the slice this pulse excites, rst note that a gradient of 0....
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This note was uploaded on 04/30/2011 for the course MECHANICAL MSC taught by Professor Dr.rahula during the Spring '10 term at University of Moratuwa.
 Spring '10
 DR.RAHULA

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